{"id":257975,"date":"2025-07-17T19:19:38","date_gmt":"2025-07-17T19:19:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=257975"},"modified":"2025-07-17T19:19:41","modified_gmt":"2025-07-17T19:19:41","slug":"calculate-the-heat-absorbed-in-joules-by-15-1-g-of-water-to-raise-its-temperature-from-22-degrees-celsius-to-51-degrees-celsius-at-constant-pressure-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/calculate-the-heat-absorbed-in-joules-by-15-1-g-of-water-to-raise-its-temperature-from-22-degrees-celsius-to-51-degrees-celsius-at-constant-pressure-2\/","title":{"rendered":"Calculate the heat absorbed (in joules) by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure"},"content":{"rendered":"\n<p>Calculate the heat absorbed (in joules) by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure). What information do we need to calculate the heat absorbed? How much heat will be absorbed to raise 15.1 g of water by one degree Celsius? However, we want to raise the temperature of our sample of water by more than 1 degree Celsius. What is the temperature change that occurs? Now calculate the heat absorbed by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure).<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the heat absorbed by water when its temperature increases, we use the formula: Q=mc\u0394TQ = mc\\Delta TQ=mc\u0394T<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>QQQ is the heat absorbed (in joules)<\/li>\n\n\n\n<li>mmm is the mass of the water (in grams)<\/li>\n\n\n\n<li>ccc is the specific heat capacity of water (4.18 J\/g\u00b0C)<\/li>\n\n\n\n<li>\u0394T\\Delta T\u0394T is the temperature change (in \u00b0C)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Information Needed:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Mass of water (m)<\/strong>: We are given that the mass of water is 15.1 g.<\/li>\n\n\n\n<li><strong>Specific heat capacity of water (c)<\/strong>: The specific heat capacity of water is 4.18\u2009J\/g\u00b0C4.18 \\, \\text{J\/g\u00b0C}4.18J\/g\u00b0C, which is the amount of heat required to raise the temperature of 1 gram of water by 1\u00b0C.<\/li>\n\n\n\n<li><strong>Temperature change (\u0394T\\Delta T\u0394T)<\/strong>: This is the difference between the final and initial temperatures. Given that the water\u2019s temperature changes from 22\u00b0C to 51\u00b0C, the change in temperature is:<\/li>\n<\/ol>\n\n\n\n<p>\u0394T=51\u00b0C\u221222\u00b0C=29\u00b0C\\Delta T = 51\u00b0C &#8211; 22\u00b0C = 29\u00b0C\u0394T=51\u00b0C\u221222\u00b0C=29\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Calculating Heat Absorbed:<\/h3>\n\n\n\n<p>Using the formula and plugging in the known values: Q=(15.1\u2009g)\u00d7(4.18\u2009J\/g\u00b0C)\u00d7(29\u00b0C)Q = (15.1 \\, \\text{g}) \\times (4.18 \\, \\text{J\/g\u00b0C}) \\times (29\u00b0C)Q=(15.1g)\u00d7(4.18J\/g\u00b0C)\u00d7(29\u00b0C) Q=15.1\u00d74.18\u00d729=1762.37\u2009JQ = 15.1 \\times 4.18 \\times 29 = 1762.37 \\, \\text{J}Q=15.1\u00d74.18\u00d729=1762.37J<\/p>\n\n\n\n<p>Thus, the heat absorbed by 15.1 g of water to raise its temperature from 22\u00b0C to 51\u00b0C is approximately <strong>1762.37 J<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Heat Absorbed for a 1\u00b0C Change:<\/h3>\n\n\n\n<p>To calculate how much heat will be absorbed to raise the temperature by 1\u00b0C: Q=m\u00d7c\u00d71\u00b0CQ = m \\times c \\times 1\u00b0CQ=m\u00d7c\u00d71\u00b0C Q=15.1\u00d74.18\u00d71=63.1\u2009JQ = 15.1 \\times 4.18 \\times 1 = 63.1 \\, \\text{J}Q=15.1\u00d74.18\u00d71=63.1J<\/p>\n\n\n\n<p>This means that to raise 15.1 g of water by 1\u00b0C, <strong>63.1 J<\/strong> of heat will be absorbed.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The temperature change that occurs is 29\u00b0C, and the total heat absorbed is 1762.37 J.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-854.jpeg\" alt=\"\" class=\"wp-image-257986\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the heat absorbed (in joules) by 15.1 g of water to raise its temperature from 22 degrees Celsius to 51 degrees Celsius (at constant pressure). What information do we need to calculate the heat absorbed? How much heat will be absorbed to raise 15.1 g of water by one degree Celsius? However, we want [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-257975","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257975","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=257975"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/257975\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=257975"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=257975"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=257975"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}